# Finite etale and choice of topology

Let $$X =$$ Spec$$(R)$$ be an affine scheme and $$f:Y \rightarrow X$$ be an affine scheme over $$X$$. Let us denote the category of affine schemes over $$X$$ as Aff$$_X$$.

In Lenstra's notes Galois Theory for Schemes (p.71 Lemma 5.10, see below) he shows that in order for $$Y$$ to be finite etale over $$X$$ it is necessary and sufficient that there exists an affine scheme $$g:W\rightarrow X$$ in Aff$$_X$$ which is surjective, flat, and finite and the pull-back $$Y\times_{X}W \cong \coprod_{N}W$$ for some finite set $$N$$.

Put another way, this theorem states (with a little more work) that $$Y$$ is finite etale over $$X$$ if and only if it is locally totally split in the fppf topology. That is, there exists an fppf coverage of $$X$$ such that $$Y$$ is totally split when pulled-back over this cover.

So we could define finite etale by saying: $$Y\rightarrow X$$ is finite etale if there exists an fppf cover of $$X$$ over which $$Y$$ is totally split. This is nice because it feels similar to the definition of a covering space from topology.

But there are (at least) two other natural choices of topology that we could put on Aff$$_X$$. Namely, the flat topology and the fpqc topology.

Question: Do each of these topologies give equivalent definitions of finite etale? i.e. is it true that:

$$Y$$ is flat locally totally split iff $$Y$$ is fpqc locally totally split iff $$Y$$ is fppf locally totally split?

Due to the relative "fineness" of these topologies we know that fppf locally totally split implies fpqc locally totally split implies flat locally totally split. I can prove that all three coincide if $$R$$ is a field. I can almost show fppf locally totally split is the same as fpqc locally totally split for an arbitrary ring; I get stuck at one step.

Any references or comments would be great! Thanks

http://websites.math.leidenuniv.nl/algebra/

Let's say that a morphism of schemes $$S' \to S$$ is "$$N$$-split" if there is an isomorphism of $$S$$-schemes $$S' \simeq \coprod_{N} S$$.

Let $$f : Y \to X$$ be a morphism of schemes and suppose that there is a faithfully flat $$W \to X$$ such that $$Y \times_{X} W \to W$$ is $$N$$-split. We show that there is an faithfully flat, finitely presented $$W' \to X$$ such that $$Y \times_{X} W' \to W'$$ is $$N$$-split.

We first note that if $$x \in X$$ is a point and $$Y \times_{X} \operatorname{Spec} \mathcal{O}_{X,x} \to \operatorname{Spec} \mathcal{O}_{X,x}$$ is $$N$$-split, then by "standard limit arguments" there is an open neighborhood $$U$$ of $$x$$ such that $$Y \times_{X} U \to U$$ is $$N$$-split. (This is good because open immersions are finitely presented, whereas the map $$\operatorname{Spec} \mathcal{O}_{X,x} \to X$$ is in general not finitely presented. This also tells us that at any point we're free to localize on $$X$$ and assume that $$R$$ is a local ring.)

If we knew that $$W \to X$$ was a filtered inverse limit of faithfully flat, finitely presented $$X$$-schemes, then we'd be done by these same limit arguments, but this is false; see this note by Bhargav Bhatt or this Stacks Project page.

So instead we'll use this cover $$W \to X$$ to deduce nice properties about $$f$$ (and the fppf cover $$W' \to X$$ we get will have little to do with $$W \to X$$). Namely, since the morphism $$f$$ is finitely presented, flat, surjective after a faithfully flat base change on $$X$$, the same is true for $$f$$ itself; see e.g. 02L0, 02L2, 0495 respectively.

Now here's a trick: we pull back $$f$$ by the fppf covering $$f$$ and replace $$f : Y \to X$$ by, say, the first projection $$p_{1} : Y \times_{X} Y \to Y$$ (and replace the cover $$W \to X$$ by $$Y \times_{X} W \to Y$$). This reduces us to the case when $$f : Y \to X$$ has a section $$s : X \to Y$$ (in our case, this comes from the diagonal embedding $$\Delta : Y \to Y \times_{X} Y$$). After pulling back by $$W \to X$$ and working Zariski locally on the base, this section $$s$$ is an open and closed immersion (more precisely, if $$S$$ is a connected scheme, any section of $$\coprod_{N} S \to S$$ is isomorphic to the inclusion of one of the $$N$$ components $$S \to \coprod_{N} S$$); the property of being an open immersion descends under faithfully flat maps (see e.g. 02L3), so $$s$$ is an open immersion. This means we may write $$Y = Y' \coprod X$$ as $$X$$-schemes, and $$Y' \to X$$ has the property that it is $$(N-1)$$-split after a faithfully flat base change, so we replace $$f$$ by $$Y' \to X$$ and conclude by downward induction on $$N$$.

• Suppose that instead of having one faithfully flat morphism $W\rightarrow X$ we instead have a (possibly infinite) family $(W_{i} \rightarrow X)_{I}$ of morphisms each of which are flat and together form a faithful family. Does this still work? We can take the coproduct of all $W_{i}$, which will surely be faithful over $X$. But is it flat? I am not sure in the case that the family is infinite --- Also, thanks for taking the time to answer my question :)
– user256340
Sep 15, 2019 at 22:24
• Yes, the coproduct will be flat; you can check using the definition and the fact that the local rings of a coproduct are those of the individual components. Sep 16, 2019 at 8:12
• Okay, great. But the coproduct will not be affine in the case that I is infinite, right?
– user256340
Sep 16, 2019 at 9:58
• Right, it will not be affine. Sep 16, 2019 at 11:06